It is important for noise control to find out how sound diffraction
is affected depending on the arrangement of noise barriers. A semi-infinite
barrier needs a heavy structure to build and if it is constructed for a house,
it should be extended long enough. It is interesting to see the difference in
diffraction when it is surrounded by a U-shaped barrier or a closed barrier. A
house has certain dimensions, however, in this paper it is discussed how sound
pressure is transferred to a single receiving point inside the barrier.

2. The half amplitude
method

If diffraction is expressed in the frequency domain only with
amplitude there is no way to have the total diffraction from each part of the
barrier. However, if the diffraction is expressed in the time domain, the
diffraction from the whole barrier can be obtained by the addition of the
effect of each part using the half amplitude method 1)2).
The
following equation expresses the sound field around a semi-infinite wedge in
Fig.1 by Biot & Tolstoy3)

Fig.1 Diffraction around a
semi-infinite wedge

The point O has the shortest path on the ridge
between the source and the receiving point.

（1）

where c is sound velocity, Ｓis a signal power and ρ
is medium density. Parameter Y and β are;

Where t is time, and

For the details of Eq.(1), see reference3).
The
equation tells us that it is the line source along the ridge giving the first
and largest contribution from the point O on the ridge along the shortest path
and that after the point the equal amplitude is given to the receiving point
simultaneously from each side of the ridge. It means that when a wedge has a
finite length, half the amplitude is lost at the first end and another half at
the other end as shown in Fig.2

Fig.2 Diffraction of a finite wedge

3. Noise barrier
arrangement

In this paper, five arrangements of thin barriers as shown in Fig.3
are discussed. (i) is a semi-infinite barrier, (ii) is a U-shaped
barrier with
two semi-infinite side barriers, (iii) is the one with two finite side
barriers, (iv) is a closed rectangular barrier and (v) is similar but
the side
barrier has a double length of (iv). The length of the barrier AB is
the same
for each arrangement, and a point source Ps1 and
a receiving point P are at the same places along the barrier AB.
Geometrical
dimensions are given in the figure.The height of the point source Ps1
is 1m and that of the receiving point P is 0.5m lower than the
height of the barrier.

Fig.3 Arrangement of barriers

The impulse response of the semi-infinite barrier is explained using
the half amplitude method in Fig.4.

Fig.4 Diffraction of the semi-infinite
thin barrier using the half amplitude method when a sound source is at Ps1.

4. Diffraction
calculation for each arrangement

(4-1) Time domain
The main interest is to find how the diffraction and the reflection
by both side barriers and the rear wall occur when they are compared with the
diffraction of the semi-infinite barrier.
The
diffraction over the barrier AD and the barrier AB in Fig.3 (ii) or
(iii) is
the first direct diffraction to the receiving point when a point source
is at Ps1. After this, the sound diffraction over the barrier AD
produces the
geometrical and the edge reflections at the barrier BC and the edge
reflection
at the barrier AB as shown in Fig.5, and similarly the barrier AB
produces
reflections at the both side barriers AD and BC.

Fig.5 Geometrical and edge reflections at the barrier BC for the diffracted wave over the barrier AD.

The
geometrical reflection for the barrier AD is calculated as the diffraction over
the barrier to the image receiving point P’ along the projection of the edge of
AD to the barrier BC, as is the diffraction over the barrier AB. In Fig.3 (iv)
and (v), the barrier CD joins to produce reflections.
The
edge reflection at the barrier BC of the diffraction over the barrier AD is
calculated as the double diffraction, though the second one is actually
reflection. First, the shortest path between Ps1 and
P’ over the edges of AD and BC is found. When the crossing point on the edge CD
is O1, and the one on the edge BC is O2, P”
is located at the extension of the path O1O2 with the distance from O2 to P’. After the
shortest path point O1, each side of the edge AD is divided into
divisions in pairs whose paths give the same arriving time to P”. Now, each
division with a half of the amplitude given by Eq. (1) is a point source for
the edge reflection at the edge BC to the receiving point P. It is calculated
by Eq. (1) as well. The contribution of each division is summed up and the edge
reflection is obtained.
The
calculations were done until the second order. The reflection of the ground was
not included. If it is necessary, the same calculation can be done for the
object in the air.
Calculated diffraction in the time domain is shown in Fig.6 for the
different barrier arrangements, when a point source is at Ps1.

Fig.6 Calculated diffraction in the time domain for the different
barrier arrangements in Fig.3 when a point source is at Ps1.

The
shortest path from the point source Ps1 to the receiving
point P for every arrangement from (ii) to (v) in Fig.3 crosses on the ridge of
the barrier AD. It crosses before the point A on the semi-infinite barrier in
Fig.3(i). Those points give the first and largest peaks in the diffraction and
the one from the semi-infinite barrier is a little less in the amplitude than
the other barriers.
The next dominant diffraction is the geometrical reflection on the
barrier BC of the diffraction over the barrier AD. At the estimation of
calculated results, the time pattern of traffic noise must be considered.
However, we can say at least that it is essential to reduce the first
diffraction over a barrier and the reflections at the other barriers.
In
this situation, if a Helmholz resonator type absorbent is used at the barrier
edge, it decreases the noise level behind it, because the line source there is
convolved with its impulse response that oscillates in positive and negative.
Therefore, it must be more efficient than porous absorbents.

(4-2) Frequency Domain
As the calculated result in the time domain has the dominant
contribution from the

Fig.7 Transfer function of the diffraction for each arrangement in
Fig.3(ii) to (v) being compared
with the one for the semi-infinite barrier

shortest path, which includes high frequency components, it is
difficult to notice how the change in the low frequency occurs because of the
low and long gradual decrease. The transfer function for each arrangement from
(ii) to (v) in Fig.3 is shown by a continuous curve in Fig.7 being compared
with the one by a dotted line for the semi-infinite barrier.
A
further important aspect is that we have to estimate the perceptive noise level
after the integration during the time window, which is assumed 40msec, of the
absolute value after a practical noise is convolved with the impulse response
of the barriers and the transient response of our hearing system4).

Discussions
When two fences meet and make a corner, the diffractions of each
fence there are different. They should be the same. This was caused by further
multiple diffractions, and they need the help of the boundary integral equation
in the time domain to be compared and find the relation.

References
1)H.Morimoto & Y.Sakurai;
”Diffraction of a semi-infinite thick barrier”, to be published on the Journal
of JASJpn (in Japanese).
2)H.Morimoto & Y.Sakurai;
”Diffraction around a rectangular body”, to be published on the Journal of
JASJpn (in Japanese).
3)I.Tolstoy; ”Wave propagation”, p.339-347, 1973, McGraw-Hill.
4)Y.Sakurai & H.Morimoto;
”Binaural hearing and time window in the transient”, J. Acoust.Soc.Jpn(E), 10,
4, p.229-233(1989).